On a test, Shoshana got 65% or 23.4 of the items correct? How many items were on the test?

Mathematics

2 answers:

bazaltina [42]3 years ago

8 0

The answer is C) 36 :) <3

Marizza181 [45]3 years ago

6 0

65 % ------------------ 23.4 items
100 % ----------------- x items
65 : 100 = 23.4 : x
65 x = 100 * 23.4
65 x = 2340
x = 2340 : 65
x = 36
Answer: there were 36 items on the test.

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zubka84 [21]

Answer:

a.10

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5 0

3 years ago

Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

inna [77]

$(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)$

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

$6 x^{6}/3x = 2 x^{5} \\ \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\$

We repeat previous steps until we run out of numbers:
$3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2$
$\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}$

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
$(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )$

6 0

3 years ago

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Vinvika [58]

Answer:

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Just place the domain values to your x value in the function.

7 0

3 years ago

Please Help!

alisha [4.7K]

Diagonal of the parallelogram divides the parallelogram in to two equal areas.

So area of parallelogram = 2(area of triangle)

According to the given diagram,

AB= 8, AD = 5 and BD = 11

So according to the Heron's formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\\\where,\\ s =\frac{a + b + c }{2}$

and a, b and c are the three sides of the triangle

Area of triangle ABD = $s=\frac{8 + 5 + 11}{2} \\\\s = \frac{24}{2}\\\\s = 12Area of triangle ABD = \sqrt{12(12 - 8) (12 - 5) (12 - 11)}\\\\Area of triangle ABD = \sqrt{12(4) (7) (1)}\\\\Area of triangle ABD = \sqrt{336}\\\\Area of triangle ABD = 18.33$